Question

Given that x is a random variable for which a Poisson probability distribution provides a good approximation, use statistical software to find the following:

a.$\mathit{P}\left(x⩽2\right)$ when $\mathit{\lambda }\mathbf{=}\mathbf{1}$

b.$\mathit{P}\left(x⩽2\right)$ when $\mathit{\lambda }\mathbf{=}\mathbf{2}$

c.$\mathit{P}\left(x⩽2\right)$ when $\mathit{\lambda }\mathbf{=}\mathbf{3}$

d. What happens to the probability of the event $\mathbf{\{}\mathit{x}\mathbf{⩽}\mathbf{2}\mathbf{\}}$ as $\mathit{\lambda }$ it increases from 1 to 3? Is this intuitively reasonable?

Short answer

a.The probability$P\left(x⩽2\right)$is obtained from the R-software is 0.6766764.

b.The probability$P\left(x⩽2\right)$is obtained from the R-software is 0.6766764.

c.The probability $P\left(x⩽2\right)$ is obtained from the R-software is 0.4231901.

d.The average probability value decreases when the $\lambda$value increases are reasonable.

Step-by-step solution

Given Information

The x is a Poisson random variable.

Sate the statistical software used to compute the probability

The R-software is used to compute the Poisson probabilities.

The general form for R-command used for computing the Poisson probability is given as follows:

$$\begin{gathered} \mathbf{dpois}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{lambda}\mathbf{,}\mathbf{}\mathbf{log}\mathbf{=}\mathbf{FALSE}\mathbf{)} \\ \mathbf{ppois}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{}\mathbf{lambda}\mathbf{,}\mathbf{}\mathbf{lower}\mathbf{.}\mathbf{tail}\mathbf{=}\mathbf{TRUE}\mathbf{)} \end{gathered}$$

State the Poisson probability distribution

The Poisson probability distribution is,

$P\left(x\right)=\frac{{\lambda }^x{e}^{-\lambda }}{x!} \left(x=0,1,2,...\right)$

The parameter is the mean of the Poisson distribution.

(a)Compute the probability  P(x⩽2)  when λ=1

The value of the Poisson parameter,$\lambda =1$ .

The$P\left(x⩽2\right)$when $\lambda =1$ can be obtained by,

$P\left(x⩽2\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)$

The R-code used to compute the$P\left(x⩽2\right)$is,

$ppois\left(2,lambda=1,lower. tail=TRUE\right)$

Therefore,

The probability $P\left(x⩽2\right)$ is obtained from the R-software is 0.9196986.

(b) Compute the probability P(x⩽2) when λ=2

The value of the Poisson parameter,$\lambda =2.$

The$P\left(x⩽2\right)$when $\lambda =2$ can be obtained by,

$P\left(x⩽2\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)$

The R-code used to compute the$P\left(x⩽2\right)$is,

$ppois\left(2,lambda=2,lower. tail=TRUE\right)$

Therefore,

The probability $P\left(x⩽2\right)$ is obtained from the R-software is 0.6766764.

(c) Compute the probability P(x⩽2) when λ=3

The value of the Poisson parameter,$\lambda =3.$

The$P\left(x⩽2\right)$when $\lambda =3$ can be obtained by,

$P\left(x⩽2\right)=P\left(x=0\right)+P\left(x=1\right)+P\left(x=2\right)$

The R-code used to compute the$P\left(x⩽2\right)$is,

$\mathbf{ppois}\left(\mathbf{2}\mathbf{,}\mathbf{lambda}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{lower}\mathbf{.} \mathbf{tail}\mathbf{=}\mathbf{TRUE}\right)$

Therefore,

The probability $P\left(x⩽2\right)$ is obtained from the R-software is 0.4231901.

(d) State the reason what happens to the probability of an event when the λvalues increase from 1 to 3

The probability of the event $\{x⩽2\}$ when $\lambda =1$ is 0.9196986.

The probability of the event $\{x⩽2\}$ when $\lambda =2$ is 0.6766764.

The probability of the event $\{x⩽2\}$ when $\lambda =3$ is 0.4231901.

Here, the event's probability decreases when the value of $\lambda$ is increases. The reason is that there are a lot more events that could happen during the interval.

Therefore, The average probability value decreases when the $\lambda$value increases are reasonable.